A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.

$\frac{3}{8}$

The correct answer is Option (3) → $\frac{3}{8}$ ##

Let $E$ be the event that the man reports that six occurs in the throwing of the die and let $S_1$ be the event that six occurs and $S_2$ be the event that six does not occur.

Then

$P(S_1) = \text{Probability that } six \text{ occurs} = \frac{1}{6}$

$P(S_2) = \text{Probability that } six \text{ does not occur} = \frac{5}{6}$

$P(E|S_1) = \text{Probability that the man reports that } six \text{ occurs when } six \text{ has actually occurred on the die}$

$= \text{Probability that the man speaks the truth} = \frac{3}{4}$

$P(E|S_2) = \text{Probability that the man reports that } six \text{ occurs when } six \text{ has not actually occurred on the die}$

$= \text{Probability that the man does not speak the truth} = 1 - \frac{3}{4} = \frac{1}{4}$

Thus, by Bayes' theorem, we get

$P(S_1|E) = \text{Probability that the report of the man that } six \text{ has occurred is actually a } six$

$= \frac{P(S_1)P(E|S_1)}{P(S_1)P(E|S_1) + P(S_2)P(E|S_2)}$

$= \frac{\frac{1}{6} \times \frac{3}{4}}{\frac{1}{6} \times \frac{3}{4} + \frac{5}{6} \times \frac{1}{4}} = \frac{1}{8} \times \frac{24}{8} = \frac{3}{8}$

Hence, the required probability is $\frac{3}{8}$.